I Kay Oh

$\large{\displaystyle I = \int^{1}_{0} \dfrac{x^3 \ln^2 (x)}{(x^4+1)^2} }dx$

Substitute $x^{4} = t$ , $\displaystyle 64I = \int^{1}_{0} \dfrac{\ln^2 (t)}{(t+1)^2}dt$

Now using $\dfrac{1}{(1+t)^{2}} = 1-2t+3t^{2}-4t^{3}+ \cdots$ $\small \color{#D61F06}{\text{(If you want proof for this,ask me in the comments section.)}}$

$\displaystyle 64I = \sum_{r=0}^{\infty} (-1)^{r} (1+r) \int_{0}^{1} t^{r} \ln^2(t) dt$

Now $\text{using the substitution } t = e^{\dfrac{-x}{r}},\displaystyle \int_{0}^{1} t^{r} \ln^2 (t) dt = \dfrac{2}{(1+r)^{3}}$

Therefore $I$ becomes $32I = \sum_{r=0}^{\infty} \dfrac{(-1)^{r}}{(r+1)^{2}}$

$\implies \displaystyle I = \dfrac{\pi^{2}}{384}$

nice problem.